5
$\begingroup$

This problem comes from the book Topology and Geometry of Bredon :

Consider the annulus. Identify antipodal points on the outer circle. Also identify antipodal points on the inner circle. Calculate the fundamental group of this surface.

Let us denote $X$ the annulus with identifications. Consider an open cover $\{U,V\}$ of the surface with $U$ and $V$ being annulus inside the big annulus where the former has its outer circle being the outer circle of the annulus with identifications, and where the latter has its inner circle being the inner circle of the annulus with identifications.

Now, the way I understand it, $U$ and $V$ strongly deformation retract to the outer and inner circle of the annulus with identifications respectively, which is $\mathbb{RP}^1$ and thus \begin{equation*} \pi_1(U) \approx \pi_1(V) \approx \pi_1(U \cap V) \approx \mathbb{Z}. \end{equation*}

By the Seifert-van Kampen's theorem, the canonical maps of the fundamental group of $U$, $V$, and $U \cap V$ into that of $X$ induce an isomorphism \begin{equation*} \pi_1(X) \approx \pi_1(U) \ast_{\pi_1(U \cap V)} \pi_1(V). \end{equation*}

I am not sure of how one could compute the free product with amalgamation? If I am not mistaken, a generator of $\pi_1(U \cap V)$ is sent to both the generator of $\pi_1(U) = \langle a \rangle$ represented by the image of the word $a^2$ and the generator of $\pi_1(V) = \langle b \rangle$ represented by the image of the word $b^2$. Therefore, the fundamental group should look like \begin{equation*} \pi_1(X) \approx \big \langle a, b \big | a^2 = b^2\big \rangle. \end{equation*}

Is this correct? If not, could someone explain me where are the flaws in the argument?

$\endgroup$
6
  • 3
    $\begingroup$ Why $a^2 = b^2 = 1$? $\endgroup$ Jan 23, 2021 at 17:56
  • 3
    $\begingroup$ Yes, that gives $a^2 = b^2$, but why did you write $a^2 = b^2 = 1$? $\endgroup$ Jan 23, 2021 at 18:58
  • 1
    $\begingroup$ I see now, I thought that the quotient introduced $a^2 = b^2 = 1$ but in reality, we quotient by the normal subgroup generated by $\phi_1(\gamma)(\phi_2(\gamma))^{-1}$, thus only $a^2b^{-2} = 1$. I have edited the post. Thank you! Is it correct now @MichaelAlbanese ? $\endgroup$
    – Rundasice
    Jan 24, 2021 at 7:03
  • 2
    $\begingroup$ @JonathanJunne Now, the group presentation is correct. $\endgroup$
    – Kevin.S
    Jan 24, 2021 at 8:03
  • 1
    $\begingroup$ For completeness: $X$ is usually called the Klein bottle. $\endgroup$ Jan 24, 2021 at 15:55

1 Answer 1

2
$\begingroup$

Your calculation of the group is correct.

Just a couple of comments though. First, instead of $\approx$, it is much more common to use $\cong$ (\cong) to denote isomorphic groups. Secondly, you said the following:

... a generator of $\pi_1(U \cap V)$ is sent to both the generator of $\pi_1(U) = \langle a \rangle$ represented by the image of the word $a^2$ and the generator of $\pi_1(V) = \langle b \rangle$ represented by the image of the word $b^2$.

The words $a^2$ and $b^2$ are not generators of $\pi_1(U)$ and $\pi_1(V)$ respectively. Instead you could say

... a generator of $\pi_1(U \cap V)$ is sent to both the word $a^2$ in $\pi_1(U) = \langle a \rangle$ and the word $b^2$ in $\pi_1(V) = \langle b \rangle$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .